91Ó°ÊÓ

Equilibrium constant (K): In the equilibrium reaction, the ratio is between the concentration of the reactant and the concentration of the product. If K's value is smaller than 1, the reaction should be moved to the left; if K is more than 1, the reaction should be moved to the right.

As

$\mathrm{Δ}G°=-RT\ln K$

And,

$K=e^{\left(-∆G°/RT\right)}$

Here, $∆\mathrm{G}°$is the change in free energy.

K is the equilibrium constant.

R is gas constant 8.314.

T is temperature.

The given reaction is:

${\mathrm{BaCl}}_2â‹\dots {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}⇌{\mathrm{BaCl}}_{2\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}$

$\mathrm{Δ\pm á}°=-63.11\mathrm{kJ}/\mathrm{mol}\mathrm{at}25°\mathrm{C}$

$\mathrm{Δ}H°=+148J(K.mol)at25°C$

The vapor pressure of the given equation is:

$K=\frac{P_{BaC{l}_2}.P_{H_{2}O}}{P_{BaC{l}_2{.}_{H_{2}O}}}$

For pure solid, the activity is equal to one.

role="math" localid="1663391640608" $$\begin{gathered} \mathbf{K}\mathbf{=}{\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}} \\ \mathbf{}\mathbf{}\mathbf{}={\mathrm{e}}^{-∆\mathrm{G}°/\mathrm{RT}} \\ ={\mathrm{e}}^{-\left(∆\mathrm{H}°-\mathrm{T}∆\mathrm{S}\right)/\mathrm{RT}} \end{gathered}$$

Here,

$$\begin{gathered} \mathrm{Δ}G°=-(\mathrm{Δ}G°-T\mathrm{Δ}S) \\ K=e^{\left\{(63.11×{10}^{3}J/mol-(298.15K\left)\right(148JK{"}^{-1})"mo{l}^{-1})/(8.314472J/(Kâ‹\dots mol\left)\right)(298.15K)\right\}} \\ K=4.7×{10}^{-4}bar \end{gathered}$$

The given reaction is:

${\mathrm{BaCl}}_2â‹\dots {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}⇌{\mathrm{BaCl}}_{2\left(\mathrm{s}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}$

$$\begin{gathered} \mathrm{Δ\pm á}°=-63.11\mathrm{kJ}/\mathrm{mol}\mathrm{at}25°\mathrm{C} \\ \mathrm{Δ\pm á}°=+148\mathrm{J}(\mathrm{K}.\mathrm{mol})\mathrm{at}25°\mathrm{C} \end{gathered}$$

$\mathrm{K}=\frac{{\mathrm{P}}_{{\mathrm{BaCl}}_2}.{\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}}{{\mathrm{P}}_{{\mathrm{BaCl}}_2}.{\mathrm{H}}_2\mathrm{O}}$

For pure solid, the activity is equal to one.

$$\begin{gathered} K=P_{H_{2}O} \\ {P}_{H_{2}O}=1=e^{-(\mathrm{Δ}H°-T\mathrm{Δ}S)/RT} \end{gathered}$$

Here,

$\mathrm{Δ\pm á}°-\mathrm{TΔ³\S }°\mathrm{must}\mathrm{be}\mathrm{zero}$

And $\mathrm{Δ\pm á}°-\mathrm{TΔ³\S }°=0$

Therefore,

$\mathrm{T}=\frac{\mathrm{Δ\pm á}°}{\mathrm{Δ³\S }°}=426\mathrm{K}=153°\mathrm{C}$